Added problems 4 and 5 and a helper function for 4 to Project Euler

ProjectEuler/Problem4.m

@@ -0,0 +1,41 @@
+%ProjectEuler/Problem4.m
+%This is a script to answer Problem 4 for Project Euler
+%Find the largest palindrome made from the product of two 3-digit numbers
+
+%Make your variables
+answer = 0;	%For the product of the two numbers
+numbers = [100:999];	%Create an array with a list of all 3 digit numbers
+palindromes = [];	%Holds all the numbers that are palindromes
+%Create 2 counters for an inner loop and an outer loop
+%This allows you to multiply 2 numbers from the same array
+outerCounter = 1;
+innerCounter = 1;
+
+while(outerCounter < size(numbers)(2))
+	innerCounter = outerCounter;	%Once you have multiplied 2 numbers there is no need to multiply them again, so skip what has already been done
+	while(innerCounter < size(numbers)(2))
+		%Multiply the two numbers
+		answer = numbers(outerCounter) * numbers(innerCounter);
+
+		%See if the number is a palindromes
+		%%WARNING - Ocatave does not have a Reverse function. I had to create one that reversed strings
+		if(num2str(answer) == Reverse(num2str(answer)))
+			%Add it to the palindromes list
+			palindromes(end + 1) = answer;
+		end
+		++innerCounter;	%Increment
+	end
+	++outerCounter;	%Increment
+end
+
+clear outerCounter;
+clear innerCounter;
+clear answer;
+clear numbers;
+
+max(palindromes)
+
+%This way is slow. I would like to find a faster way
+%{
+The palindrome can be written as: abccba Which then simpifies to: 100000a + 10000b + 1000c + 100c + 10b + a And then: 100001a + 10010b + 1100c Factoring out 11, you get: 11(9091a + 910b + 100c) So the palindrome must be divisible by 11. Seeing as 11 is prime, at least one of the numbers must be divisible by 11
+%}